Obviously there must be 2 $5$s So it can be easily seen that $900= 9 \cdot 4 \cdot 5 \cdot 5 \cdot 1 = 9 \cdot 2 \cdot 5 \cdot 5 \cdot 2 = 6 \cdot 6 \cdot 5 \cdot 5 \cdot 1 =6 \cdot 2 \cdot 5 \cdot 5 \cdot 3$ Conclusion follows

There are five ways of expresing $900$ in this matter, $900=1\cdot 4\cdot 9\cdot 5\cdot 5$, $900=1\cdot 6\cdot 6\cdot 5\cdot 5$, $900=2\cdot 2\cdot 9\cdot 5\cdot 5$, $900=3\cdot 3\cdot 4\cdot 5\cdot 5$, $900=2\cdot 3\cdot 6\cdot 5\cdot 5$. So the answer is $5\cdot 5!=\boxed {600}$ numbers.

Com10atorics wrote: There are five ways of expresing $900$ in this matter, $900=1\cdot 4\cdot 9\cdot 5\cdot 5$, $900=1\cdot 6\cdot 6\cdot 5\cdot 5$, $900=2\cdot 2\cdot 9\cdot 5\cdot 5$, $900=3\cdot 3\cdot 4\cdot 5\cdot 5$, $900=2\cdot 3\cdot 6\cdot 5\cdot 5$. So the answer is $5\cdot 5!=\boxed {600}$ numbers. You overcounted all the cases. Here is the correct solution-- $2\cdot\frac{5!}{2!} + 3\cdot\frac{5!}{2! . 2!} =210$

electrovector wrote: Obviously there must be 2 $5$s So it can be easily seen that $900= 9 \cdot 4 \cdot 5 \cdot 5 \cdot 1 = 9 \cdot 2 \cdot 5 \cdot 5 \cdot 2 = 6 \cdot 6 \cdot 5 \cdot 5 \cdot 1 =6 \cdot 2 \cdot 5 \cdot 5 \cdot 3$ Conclusion follows You forgot 4,3,3

$900$ can be expressed as a product of 5 numbers: $$900=1\cdot 4\cdot 5\cdot 5\cdot 9$$$$900=2\cdot 3\cdot 5\cdot 5\cdot 6$$$$900=2\cdot 2\cdot 5\cdot 5\cdot 9$$$$900=3\cdot 3\cdot 4\cdot 5\cdot 5$$$$900=1\cdot 5\cdot 5\cdot 6\cdot 6$$the first one has 60, the second one has 60, the third one has 30, the fourth one has 30 and the last one has 30, in conclusion $210$ is the correct answer :v because we are counting numbers LOL (mistakes could appear if we take $5! \cdot 5$), every digit is less than 10

$900=2^2\times3^2\times5^2$ $900$ have 6 prime divisors, but I have to show $900$ as product of five integers. So I have to "combine" 2 prime divisors ($2\times3=6$) $23556$ and its permutations$\Rightarrow\frac{5!}{2!}=60$ $15566$ and its permutations$\Rightarrow\frac{5!}{2!2!}=30$ $33455$ and its permutations$\Rightarrow\frac{5!}{2!2!}=30$ $22559$ and its permutations$\Rightarrow\frac{5!}{2!2!}=30$ $14559$ and its permutations$\Rightarrow\frac{5!}{2!}=60$ $$60+30+30+30+60=210$$